Month: May 2016

# 127 Accelerated Tied Buoy

When stationary, the upthrust must balance the tension in the rubber band plus the weight of the ping pong ball.

ρVg = T + mg

When the bucket is accelerated upward at a, the apparent g of the water becomes g+a. The water thus exerts a larger upthrust on the ball than before. Applying N2L on the ball,

ρV(g+a) – Tmg = ma’

ρVa = ma’

a’=ρV/m a

Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket accelerates at a, the ping pong ball accelerates at >a, and the rubber becomes more stretched.

On the other hand, when the bucket has a downward acceleration of a, the apparent g of the water becomes g-a. The water exert a smaller upthrust on the ball than before. Applying N2L on the ball,

T + mgρV(ga) – = ma’

ρVa = ma’

a’=ρV/m a

Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket has a downward acceleration of a, the ping pong ball has a downward acceleration >a, and the rubber becomes less stretched.

The simplest case to consider is if the bucket is released from rest. As the bucket free falls, the upthrust acting on the ball disappears. While the bucket and the water free fall at g, the ball accelerates at larger than g, because it would be pulled down by gravitational force plus the tension in the rubber band.

# 126 Artificial g: Buoy in Water Pail

Whether stationary or in circular motion, the ball displaces the same amount of water. Yet, the upthrust acting on the ball varies continuously during the circular motion to provide the required centripetal force.

Consider when the bucket is at the bottom most position.

If the bucket is stationary, then

1. A column of water of height h would have weight hAρg. This weight must be supported by the upward pressure force exerted by the water below. Hence the water pressure at depth h must be hρg.
2. The pressure gradient results in the ball experiencing an upthrust U = ρVg, where V is the volume of displaced water.
3. ρVgmg = 0. Hence the ball is stationary.

If the bucket is in circular motion, then

1. A column of water of height h would have weight hAρg. In addition to supporting this weight, the upward pressure forces exerted by the water below must also provide for the required centripetal force of hAρ(2). Hence the water pressure at depth h must be hρ(2+g).
2. The larger pressure gradient results in the ball experiencing a larger upthrust U = ρV((2+g).
3. The net force experienced by the ball is thus ρV(2+g) –mg = ρV2 = mrω2. Hence the ball does the same circular motion as the bucket of water.

When the bucket is in circular motion, the water pressure becomes larger because the water molecules would be pressing harder into one another. This is equivalent to the water experiencing an artificial g that is directed in the centrifugal direction, and an “upthrust” that is always in the centripetal direction. The magnitude of this artificial g would be varying continuously (in the exact manner as the tension in the rope, and normal contact force in the bucket) during the circular motion, so as to keep the water and the ball in circular motion.

# 125 Water in Pail Swung in a Vertical Circle

Why doesn’t the water fall out of the bottle?

Let’s consider the water when the bottle is at the top most position. At this instant, the water has a horizontal velocity. The earth’s gravitational pull would have caused the water to continue along a parabolic arc. The bottle, however, intends to follow a circular path. Relative to the circular path of the bottle, the parabolic path of the water would have taken the water out of the bottle through the TOP, not the BOTTOM, thus puncturing the base of the bottle.

Of course, the bottle would not allow itself to be punctured by water. So what it does is to exert a downward normal contact force on the water, just enough to push the water into following the same circular path as the bottle.

The water falls out of the bottle only if it falls faster than the pail. As long as the bottle is swung at a high enough speed, this will not happen. In fact, the bottle must press down on the water to make the water fall faster, as fast as the bottle.

# 124 Ball Bearing in Circles

The wall of the container pushes the ball bearing in the centripetal direction, causing the ball bearing’s velocity to keep changing its direction. When the centripetal force dissappears (when container is lifted), the ball bearing continues at its current velocity (as according to Newton’s 1st Law), which is in the tangential direction.

# 123 Circling Buoys

When the bottles are going round circles, the water level in the bottle is not level anymore. Instead, water level decreases from the outside to the inside.

The ping pong ball that is submerged in water experience pressure forces due to water. Since the pressure is always larger on the outside compared to inside, the ping pong ball experiences a net pressure force towards the inside.

In other words, because the pressure of the water increases not just with depth, but also towards the outside, the ping pong experiences an additional “insidethrust” on top of the usual upthrust. It is this “insidethrust” (minus the horizontal component of tension) that provides the required for the ping pong ball to undergo circular motion.